Mastering Mathematics Smartly by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Scalar product

Students typically find the topic of "Vectors" difficult.

Let us simplify its complexity by considering the applications of the scalar product.

We use the scalar product in these ways:

(1) To find the angle between 2 vectors.

(2) To find the angle between 2 lines.

(3) To find the length of projection between of a vector onto another.

(4) To find the angle between 2 planes.

(5) To find the angle between a line and a plane.

(6) To find the distance of the plane from the origin.

See summary.

Always represent the problem description with any of the diagrams in the summary, to help you solve the question.

Evaluating arbitrary scalar products

We look at a type of question that used to be popular in the past, but has yet been asked in any H2 maths exam. This sort of question involves arbitrary vectors (i.e. vectors with no specific numerical values given). Invest time to study this kind of problem carefully.

Question:
With reference to the origin O, the position vectors A and B are a and b respectively. The point C lies on AB produced such that AB = 3BC. Find an expression for the position vector of N, the foot of the perpendicular from C to the line OB.

Solving approach:
1. Find the position vector of C, using the fact that AB = 3BC.

\textbf{b} - \textbf{a} = 3(\textbf{c} - \textbf{b})
\Rightarrow \textbf{c} = \frac{1}{3}(4\textbf{b} - \textbf{a}).

2. Find the equation of the line OB, on which N lies.

l_{OB} : \textbf{r} = \lambda\textbf{b}, \lambda \in \mathbb{R}.
Since N lies on this line, \overrightarrow{ON} = \lambda\textbf{b} for some fixed \lambda (which is to be found in the next step).

3. Find \lambda, using the fact that CN is perpendicular to the line OB.

\overrightarrow{CN} = \left(\lambda - \frac{4}{3}\right)\textbf{b} + \frac{1}{3}\textbf{a}.

\overrightarrow{CN}.\textbf{b} = 0

\Rightarrow \left(\lambda - \frac{4}{3}\right)\textbf{b}.\textbf{b} + \frac{1}{3}\textbf{a}.\textbf{b} = 0

\Rightarrow \lambda = \frac{4}{3} - \frac{\textbf{a}.\textbf{b}}{3\textbf{b}.\textbf{b}}.

4. Obtain the position vector of N.

\overrightarrow{ON} = \left(\frac{4}{3} - \frac{\textbf{a}.\textbf{b}}{3\textbf{b}.\textbf{b}}\right)\textbf{b}.

One should realise that the steps are exactly identical to the situation where vectors are given numerical values.

Proceed to the discussion on arbitrary vector products.